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Find the vertex of the parabola y = -5x² - 6x + 15/2

Sagot :

[tex]\\ \sf\longmapsto y=-5x^2-6x+15/2[/tex]

[tex]\\ \sf\longmapsto -5x^2-6x-y+15/2=0[/tex]

  • Multiply 2 with each term

[tex]\\ \sf\longmapsto -10x^2-12x-2y+15=0[/tex]

  • Convert to vertex form

[tex]\\ \sf\longmapsto y=-5(x^2+\dfrac{6}{5}x+\dfrac{9}{25})+\dfrac{93}{10}[/tex]

[tex]\\ \sf\longmapsto y=-5(x^2+2\dfrac{3}{5}x+(\dfrac{3}{5})^2)+\dfrac{93}{10}[/tex]

[tex]\\ \sf\longmapsto y=-5(x+\dfrac{3}{5})^2+\dfrac{93}{10}[/tex]

Compare to vertex form of parabola

[tex]\boxed{\sf y=a(x-h)^2+k}[/tex]

Now

  • h=-3/5
  • k=93/10
Mhanifaidn

Answer:

  • (- 0.6, 9.3)

Step-by-step explanation:

Given parabola:

  • y = -5x² - 6x + 15/2

The vertex is determined by x = - b/2a:

  • x = - (-6)/(2*(-5)) = -0.6

Find y- coordinate of vertex:

  • y = - 5(0.6)² - 6(- 0.6) + 15/2 = 9.3
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